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Reaction scheme for diastereoselective reduction of C=O

In the reactions above, taken from Tetrahedron Lett. 1982, 23 (23), 2355–2358, the choice of protecting group on the α-hydroxyl group has a large influence on the diastereoselectivity of reduction.

Why does the benzyloxymethyl protecting group lead to the formation of the anti diol (consistent wiht the Cram chelate model), whereas the tert-butyldiphenylsilyl protecting group gives the syn diol (consistent with the Felkin–Anh model)?

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In order for the Cram chelate product to predominate after the addition of hydride donor to a chiral carbonyl compound, which contains a heteroatom in the alpha position, this heteroatom and part of the reagent must be able to form a five-membered ring chelate. If this is not possible, one observes Felkin-Anh proudct.

In the mentioned reaction, when protecting group is not bulky it form five-membered ring chelate easily with Li+ after addition of Hydride to carbonyl resulted in cram chelate product as major product. However, when bulky protecting group is present it doesn't five-membered ring, consequently resulted in Felkin-Anh product as major product.

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I think Jan did an excellent job of answering your question, however I am writing a slightly longer answer largely with more details concerning the Felkin–Anh rule, just so that someone who is not already familiar with them can learn something too.

Assuming that there isn’t an unusually electronegative atom on the carbon next to the carbonyl, the largest group prefers a conformation where it is perpendicular to the carbonyl moiety. This gives two relatively competitive lowest energy conformations, with the medium and smallest groups differing in their proximity to the carbonyl oxygen.

For the TBDPS-protected substrate shown above, the OTBDPS group would be considered the largest group. The medium and small groups are methyl and hydrogen respectively. This placement is also consistent with a modification to the Felkin–Anh model, which places electronegative substituents (such as –NR2 or –OR) perpendicular to the carbonyl (an explanation is given at the bottom of the answer). In any case, the two conformations of interest are the two shown below:

Reactive conformations of TBDPS-protected substrate

Moreover, the nucleophile doesn't prefer to attack the carbonyl carbon at a 90° angle; rather it prefers something closer to the tetrahedral angle (Bürgi–Dunitz trajectory). Naturally, the nucleophile prefers attack from the opposite side to the large group. This leaves two possibilities: for the nucleophile to attack over the small group, or for it to attack nearest to the medium group. The former is preferred, and the major diastereomer will be the syn diol:

Formation of Felkin–Anh diastereomer

On the other hand, Lewis basic groups such as $\ce{-OCH2OBn}$ are also capable of forming a chelate with the metal together with the carbonyl oxygen. This feature forces the $\ce{OR}$ group to be held parallel to the carbonyl. Nucleophilic attack still occurs along the Bürgi–Dunitz trajectory, and the nucleophile attacks opposite the larger of the two remaining groups. This will invert the stereochemical outcome as compared to the Felkin–Anh model previously described, and hence leads to the formation of the anti diol.

Cram chelate model for BOM -protected substrate

Now, to answer your question of why does the choice of protecting group matter. The previous discussion illustrates that the reason we get the Cram Chelate product is the chelation of the metal by the $\ce{OR}$ oxygen and the carbonyl. This chelation can take place easily if the $\ce{R}$ group is small, but is greatly hindered with a bulky alkyl/silyl substituent, in which case we get the product determined by the polar Felkin–Anh relationship.

The bulk of the TBDPS group does indeed prevent chelation, and you get the observed stereochemistry in accord with the Felkin–Anh rule. However, chelation is possible when BOM is the protecting group (Jan already offered an excellent explanation), and you end up with the Cram chelate product.


When there is an electronegative atom or group α to the carbonyl, there is a slight change to the Felkin–Anh model which may be applied. This ‘slight change’ is that this electronegative group becomes the equivalent of the ‘large’ group in the Felkin–Anh model.

The reason for this that the energy of the C–X σ* antibonding orbital is rather low, and so when the C–X bond is perpendicular to the carbonyl group, this can overlap with the π* MO of the carbonyl group to make a new, lower energy LUMO.

In "human" speak, this means that this conformational arrangement is more reactive than any other. With this change in mind, the Felkin–Anh type addition now proceeds as before. However, the stereochemical outcome will generally be different from the original Felkin–Anh model.

For more information, see: How is the Felkin-Anh model of stereoinduction correctly explained with MO theory?

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The $\ce{TBDPS}$ group (tert-butyldiphenylsilyl) is a very bulky protecting group. A silicon atom is bonded to the oxygen, which already acts as a larger carbon atom. That is then substituted by three alkyl or aryl groups, one of which is the very bulky tert-butyl group. There is a lot of congestion going on next to that oxygen atom, so its ability to coordinate to Lewis acids (including via hydrogen bonds) is severely limited. Therefore, it is mainly the polar Felkin–Anh relationship that dictates the stereochemistry of the attack in the case of a $\ce{TBDPS}$ group.

On the contrary, the benzyloxymethyl (BOM) group $\ce{CH2OBn}$ used in the other example has a primary carbon attached directly to the protected oxygen. Not only is carbon a lot smaller than silicon, but it also features two hydrogen substituents ($\ce{CH2}$) and only one substituent group – an oxygen which itself only bears one other group. Therefore, the BOM-protected oxygen can still coordinate to Lewis acids easily, such as the lithium cation or aluminium as present in the example. Thus, the Cram chelate model is feasible in this case and correctly predicts the observed stereochemistry.


This is quite often the case that some protecting groups allow for chelation while others don’t. Pretty much all the silicon-based protecting groups (those that include a $\ce{O-Si}$ bond) will inhibit chelation due to their size. Typically anything that starts off as $\ce{CH2}$ will allow chelation. A non-exhaustive list of common groups of this type would be:

  • MOM
  • SEM
  • Bn
  • PMB
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